The Last Number

Use iduction, and we have this formula;

$\zeta=\sum_{i=1}^ni^3=({ \frac{n(n+1)}{2} })^2.$

So we can,

$({ \frac{n(n+1)}{2} })^2 <250000 ...(1)$

$\Leftrightarrow{ \frac{n(n+1)}{2} }<500$

$\Leftrightarrow{ {n(n+1)} }<1000$

For $n=31$ we have $31*32=992 <1000.$ Substitute $n=31$ to L.H.S $...(1)$

Because $({ \frac{31(31+1)}{2} })^2=246016,$ so we take the next number $n=32,$ $n=32 \implies n^2=32^2= \boxed{1024}.$

Now,the diference $250000$ from $246016$ are $3984$ ,

$\lfloor{ \frac{3984}{32^2}} \rfloor = \boxed{3}$ times.

the sequence can be simplified to sum(n^3) which gives n(max)=31. so, 961 961s are included in this sum. the remaining sum is 3986. the next square is 1024. this means that 3 1024s more can fit in the sum. Hope I made sense!

I probably have to prove this, but I solved using observation. $1^{2} = 1^{2}. T_{1}$ $1^{2} + 2^{2} + 2^{2} = 3^{2}. T_{2}$ $1^{2} + 2^{2} + 2^{2}+ 3^{3} + 3^{3} + 3^{3} = 6^{2}. T_{3}$ Notice how the second column, all the numbers that are squared are triangular numbers, given by $T_{n} = \frac{n(n+1)}{2}$

The max first column value is 250000, or $500^{2}$ . I set $500 = \frac{n(n+1)}{2}$ and the nearest positive integer you get from solving for n is 31. Evaluating at $T_{31}$ you get 246016. Subtract 246016 from 250000. Then, you know that the next term up is term 32. But, you know that you will go over 250000 if you add $32\cdot 32^{2}$ , so just count the maximum number of $32^{2}$ that have to be subtracted from $250000-246016$ so that you don't get a negative number. That happens to be subtracting $32^{2}$ thrice.